A man, in search of a wife, tries two approaches: generous and cheapskate. When

Question

A man, in search of a wife, tries two approaches: generous and cheapskate. When the man tries the generous approach, he ends up spending $1,000 on his date, who will, eventually, with probability 0.95 break up with him, but with probability 0.05 marry him. When the man tries the cheapskate approach, he spends $50 on his date, who will eventually break up with him. So far in his life, the man has only experienced failure, so he cannot tell which approach works better. He therefore decides to choose an approach (generous or cheapskate) at random. (a) Assuming the man starts searching today, what is his expected cost to find a wife? (b) Compute the variance on the amount of money the man ends up spending to find a wife. A man, in search of a wife, tries two approaches: generous and cheapskate. When the man tries the generous approach, he ends up spending $1,000 on his date, who will, eventually, with probability 0.95 break up with him, but with probability 0.05 marry him. When the man tries the cheapskate approach, he spends $50 on his date, who will eventually break up with him. So far in his life, the man has only experienced failure, so he cannot tell which approach works better. He therefore decides to choose an approach (generous or cheapskate) at random. (a) Assuming the man starts searching today, what is his expected cost to find a wife? (b) Compute the variance on the amount of money the man ends up spending to find a wife.

Explanation / Answer

Solution

Let A and B represent the two approaches, generous and cheapskate respectively and the respective dating costs be C1 and C2.

Back-up Theory

If a discrete random variable, X, has probability function p(x), then

Mean (average) of X = E(X) = sum{x.p(x)} summed over all possible values of x…..…. (1)

E(X2) = sum{(x2).p(x)} summed over all possible values of x…………………………....(2)

Variance of X = V(X) = E(X2) – { E(X)}2………………………………………………..(3)

Random selection => equal probability ……………………………………………………(4)

Now, to work out the solution,

Part (a)

Given ‘He therefore decides to choose an approach (generous or cheapskate) at random.’, by (4) above, P(A) = P(B) = ½ ………………………………………………………………..(5)

Let C be the man’s cost to find a wife, then by (1) and (5) above, expected cost to find

a wife = E(C) = {C1 x P(A)} + {C2 x P(B)}

= (1000 x ½) + (50 x ½) [given C1 = 1000 and C2 = 50]

= $525 ANSWER

Part (b)

Variance on the amount of money the man ends up spending to find a wife = V(C).

Now, E(C2) = {C12 x P(A)} + {C22 x P(B)} [vide (2) above]

= (½)(10002 + 502)

= 501250

Vide (3) above,

V(C) = 501250 - 5252

= 225625 ANSWER

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